Question

A particle is projected with a speed $V$ from a point O making an angle of ${30}^{\circ }$ with the vertical. At the same instant, a second particle is thrown vertically upwards with a speed $v$ from a point A. The two particles reach $H$, the highest point of the parabolic path of particle one simultaneously. Then ratio $V/v$ is

• A. $3\sqrt{2}$
• B. $2\sqrt{3}$
• C. $\frac{2}{\sqrt{3}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $\frac{2}{\sqrt{3}}$

The maximum height reached by the particle 1 will be
$H=\frac{(Vsin{60}^{\circ }{)}^2}{2g}.......\left(1\right)$
The maximum height reached by the particle 2 will be
$H=\frac{v^2}{2g}.......\left(2\right)$
By (1) & (2)
$\frac{V^2}{v^2}{\left(\frac{\sqrt{3}}{2}\right)}^2=1$
$\frac{V}{v}=\frac{2}{\sqrt{3}}$